Problem 6 Find the area under the graph of... [FREE SOLUTION]

Chapter 4: Problem 6

Find the area under the graph of $\int$ over the interval [-6,4] $$f(x)=\left\\{\begin{array}{lll} -x-1, & \text { for } & x < -1 \\ -x^{2}+4 x+5, & \text { for } & x \geq-1 \end{array}\right.$$

Short Answer

Expert verified

The area under the graph is approximately 68.83.

Step by step solution

- Identify the bounds and the function

The exercise asks for the area under the graph of a piecewise function over \[-6, 4\]. The function is given by: $f(x)=\begin{cases} -x-1, & \text{for} & x < -1 \ -x^2 + 4x + 5, & \text{for} & x \geq -1 \end{cases} $.

- Split the integral at the piecewise point

Since the function changes at $x = -1$, we need to split the integral at this point. This gives us: \[ \text{Area} = \int_{-6}^{-1} (-x-1) \, dx + \int_{-1}^{4} (-x^2 + 4x + 5) \, dx \]

- Integrate the first piece

Integrate $ -x-1 $ from $-6$ to $-1$: \[\int_{-6}^{-1} (-x-1) \, dx = \left[ -\frac{x^2}{2} - x \right]_{-6}^{-1} \= \left( -\frac{(-1)^2}{2} - (-1) \right) - \left( -\frac{(-6)^2}{2} - (-6) \right) \= \left( -\frac{1}{2} + 1 \right) - \left( -18 + 6 \right) \= \left( \frac{1}{2} \right) - (-12) = \frac{1}{2} + 12 = 12.5 \]

- Integrate the second piece

Integrate $ -x^2 + 4x + 5 $ from $-1$ to $4$: \[\int_{-1}^{4} (-x^2 + 4x + 5) \, dx = \left[ -\frac{x^3}{3} + 2x^2 + 5x \right]_{-1}^{4} \= \left( -\frac{4^3}{3} + 2 \cdot 4^2 + 5 \cdot 4 \right) - \left( -\frac{(-1)^3}{3} + 2 \cdot (-1)^2 + 5 \cdot (-1) \right) \= \left( -\frac{64}{3} + 32 + 20 \right) - \left( \frac{1}{3} + 2 - 5 \right) \= \left(\frac{-64}{3} + 52\right) - \left( -\frac{2}{3} \right) \= \left(\frac{-64 + 156}{3}\right) + \frac{2}{3} = \frac{94}{3}\]

- Sum up the results

Add the results of the integrals from steps 3 and 4 to get the total area: \[ \text{Total Area} = 12.5 + \frac{94}{3} = \frac{37.5}{1} + \frac{94}{3} = \frac{112.5}{3} + \frac{94}{3} = \frac{206.5}{3} = 68.83 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise Functions

A piecewise function is defined by different expressions depending on the value of the input variable. This type of function is useful when a situation calls for different rules or equations in different segments of the domain.

The given function is piecewise because it has different formulas for different intervals of x.
For x < -1, the function is defined as $f(x) = -x - 1$.
For x 鈮� -1, the function is defined as $ f(x) = -x^2 + 4x + 5$.

These changes in the structure of the function are usually marked clearly within the function鈥檚 definition. In our example, the transition occurs at x = -1. When solving integrals that involve piecewise functions, it鈥檚 important to treat each part separately and then combine the results.

Integration

Integration is a fundamental concept in calculus that helps find the total accumulation of quantities. When you integrate a function, you're essentially finding the area under the curve of the function between two points on the x-axis.

For our exercise, to find the total area under the curve, we need to perform integration twice because of the piecewise nature of the function.
The integral splits at x = -1, giving us two separate integrals to solve:

\[ \text{Area} = \int_{-6}^{-1} (-x-1) \, dx \]
and
\[ \int_{-1}^{4} (-x^2 + 4x + 5) \, dx \]

The results of these integrations provide the areas for their respective intervals. Summing these results gives the total area under the function over the interval [-6, 4].

Area Under a Curve

Finding the area under a curve is a common problem in calculus that can describe various physical concepts such as distance, area, and total accumulation.

The area under a function between two points can be found using definite integrals.
For piecewise functions, the total area is found by calculating the area under each segment and adding them together.

In this exercise, by integrating $-x - 1$, we find the area under the curve from -6 to -1. Similarly, by integrating $-x^2 + 4x + 5$ from -1 to 4, we find the corresponding area. Adding these two areas together provides the total area under the curve for the given piecewise function over the interval [-6, 4]. This final result represents our total integrated value, giving a complete sense of the area encompassed by the function鈥檚 graph.

91影视

Find the area under the graph of \(\int\) over the interval [-6,4] $$f(x)=\left\\{\begin{array}{lll} -x-1, & \text { for } & x < -1 \\ -x^{2}+4 x+5, & \text { for } & x \geq-1 \end{array}\right.$$

Short Answer

Step by step solution

- Identify the bounds and the function

- Split the integral at the piecewise point

- Integrate the first piece

- Integrate the second piece

- Sum up the results

Key Concepts

Piecewise Functions

Integration

Area Under a Curve

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